Hi everyone. I think u got used of my stupid questions...sorry for that.
If the current is still flowing through the circuit when the pedal is off, shouldn't the battery die very fast? or this energy is just trapped in the circuit and not dissipated...
:P now im regrating escaping the physics lessons to chase some chicks....ehehhe
No, the circuit is still running.
All you are doing with the footswitch is redirecting your input signal to avoid the circuit all together.
Try this:
Get a long delay, aqua boy delux or a Dirtbag.
Play something and make the repeats go to near self-oscillation.
click the footswith.
you've bypassed the signal.
play your guitar, and nothing will happen. The signal goes straight through.
Now click the switch again. The repeats are still there because the circuit never stopped.
Quote from: icecycle66 on November 07, 2013, 07:45:55 PM
No, the circuit is still running.
All you are doing with the footswitch is redirecting your input signal to avoid the circuit all together.
Try this:
Get a long delay, aqua boy delux or a Dirtbag.
Play something and make the repeats go to near self-oscillation.
click the footswith.
you've bypassed the signal.
play your guitar, and nothing will happen. The signal goes straight through.
Now click the switch again. The repeats are still there because the circuit never stopped.
I understand this. but if is still running, the battery would die soon, right?
Quote from: Gledison on November 07, 2013, 07:50:44 PM
Quote from: icecycle66 on November 07, 2013, 07:45:55 PM
No, the circuit is still running.
All you are doing with the footswitch is redirecting your input signal to avoid the circuit all together.
Try this:
Get a long delay, aqua boy delux or a Dirtbag.
Play something and make the repeats go to near self-oscillation.
click the footswith.
you've bypassed the signal.
play your guitar, and nothing will happen. The signal goes straight through.
Now click the switch again. The repeats are still there because the circuit never stopped.
I understand this. but if is still running, the battery would die soon, right?
The battery would die in no more time than the effect active and running...plus the current draw of the LED (20mA) and perhaps a tiny bit from the guitar signal working the amps.
That is why battery powered effects need you to unplug the cord to actually shut them off.
It is possible to turn the power to the circuit off when bypassed but it would likely result in a huge "thud" as the components get power. This also would make the switching more complex and also perhaps a delay while you wait for it to boot up.
Yes, the circuits are always running. The resason is that it would create a huge pop everytime you switch it on (or off).
Usually a TRS input is used to switch the power. So when using batteries always unplug the input.
Don't use batteries anyway.
Edit: Sorry. Only now I realized that rullywowr already said it all.
Quote from: kothoma on November 08, 2013, 06:45:22 AM
Yes, the circuits are always running. The resason is that it would create a huge pop everytime you switch it on (or off).
Usually a TRS input is used to switch the power. So when using batteries always unplug the input.
Don't use batteries anyway.
Edit: Sorry. Only now I realized that rullywowr already said it all.
Thanks for the info, and yes, I want to use as less as possible batteries..:)
what about when is plugged with the power supply? than even unplugging the guitar cable, the current still running trhough the circuit...any way of cutting without removing the wall wart from the wall? cheers
Quote from: Gledison on November 08, 2013, 08:53:42 AM
Quote from: kothoma on November 08, 2013, 06:45:22 AM
Yes, the circuits are always running. The resason is that it would create a huge pop everytime you switch it on (or off).
Usually a TRS input is used to switch the power. So when using batteries always unplug the input.
Don't use batteries anyway.
Edit: Sorry. Only now I realized that rullywowr already said it all.
Thanks for the info, and yes, I want to use as less as possible batteries..:)
what about when is plugged with the power supply? than even unplugging the guitar cable, the current still running trhough the circuit...any way of cutting without removing the wall wart from the wall? cheers
It depends on your wiring of the DC jack in conjunction with the battery.
It is possible to switch both with a TRS (tip ring sleeve) type ("stereo") input jack.
If you insert the typical TS guitar cable plug ("mono") then you bridge Ring and Sleeve.
This is used to switch power. Nice trick.
Quote from: kothoma on November 08, 2013, 09:19:43 AM
Quote from: Gledison on November 08, 2013, 08:53:42 AM
Quote from: kothoma on November 08, 2013, 06:45:22 AM
Yes, the circuits are always running. The resason is that it would create a huge pop everytime you switch it on (or off).
Usually a TRS input is used to switch the power. So when using batteries always unplug the input.
Don't use batteries anyway.
Edit: Sorry. Only now I realized that rullywowr already said it all.
Thanks for the info, and yes, I want to use as less as possible batteries..:)
what about when is plugged with the power supply? than even unplugging the guitar cable, the current still running trhough the circuit...any way of cutting without removing the wall wart from the wall? cheers
It depends on your wiring of the DC jack in conjunction with the battery.
It is possible to switch both with a TRS (tip ring sleeve) type ("stereo") input jack.
If you insert the typical TS guitar cable plug ("mono") then you bridge Ring and Sleeve.
This is used to switch power. Nice trick.
Hey mate. Im following the offboard wiring diagram from madbean. I do have a stereo input jack. So, 1 positive wire goes from the DC jack to the PCB and the other positive to the LED-1st lug on 3PDT and to the battery snap...
am I doing something wrong?
cheers
Quote from: Gledison on November 08, 2013, 09:32:38 AM
Quote from: kothoma on November 08, 2013, 09:19:43 AM
Quote from: Gledison on November 08, 2013, 08:53:42 AM
Quote from: kothoma on November 08, 2013, 06:45:22 AM
Yes, the circuits are always running. The resason is that it would create a huge pop everytime you switch it on (or off).
Usually a TRS input is used to switch the power. So when using batteries always unplug the input.
Don't use batteries anyway.
Edit: Sorry. Only now I realized that rullywowr already said it all.
Thanks for the info, and yes, I want to use as less as possible batteries..:)
what about when is plugged with the power supply? than even unplugging the guitar cable, the current still running trhough the circuit...any way of cutting without removing the wall wart from the wall? cheers
It depends on your wiring of the DC jack in conjunction with the battery.
It is possible to switch both with a TRS (tip ring sleeve) type ("stereo") input jack.
If you insert the typical TS guitar cable plug ("mono") then you bridge Ring and Sleeve.
This is used to switch power. Nice trick.
Hey mate. Im following the offboard wiring diagram from madbean. I do have a stereo input jack. So, 1 positive wire goes from the DC jack to the PCB and the other positive to the LED-1st lug on 3PDT and to the battery snap...
am I doing something wrong?
cheers
Nothing wrong with that per se. But this standard wiring (http://www.madbeanpedals.com/tutorials/downloads/StandardWiring_MBP.pdf) only switches the battery with the input jack. And this is fine for most folks. Because who likes to unplug all boxes on a pedal board if you can simply switch off the power supply?
Psst. But if you really want to also switch the DC jack, simply change
the wire going from the DC jack to the sleeve of the input jack so that it goes to the ring like the black wire of the battery clip.
Quote from: kothoma on November 08, 2013, 09:44:22 AM
Quote from: Gledison on November 08, 2013, 09:32:38 AM
Quote from: kothoma on November 08, 2013, 09:19:43 AM
Quote from: Gledison on November 08, 2013, 08:53:42 AM
Quote from: kothoma on November 08, 2013, 06:45:22 AM
Yes, the circuits are always running. The resason is that it would create a huge pop everytime you switch it on (or off).
Usually a TRS input is used to switch the power. So when using batteries always unplug the input.
Don't use batteries anyway.
Edit: Sorry. Only now I realized that rullywowr already said it all.
Thanks for the info, and yes, I want to use as less as possible batteries..:)
what about when is plugged with the power supply? than even unplugging the guitar cable, the current still running trhough the circuit...any way of cutting without removing the wall wart from the wall? cheers
It depends on your wiring of the DC jack in conjunction with the battery.
It is possible to switch both with a TRS (tip ring sleeve) type ("stereo") input jack.
If you insert the typical TS guitar cable plug ("mono") then you bridge Ring and Sleeve.
This is used to switch power. Nice trick.
Hey mate. Im following the offboard wiring diagram from madbean. I do have a stereo input jack. So, 1 positive wire goes from the DC jack to the PCB and the other positive to the LED-1st lug on 3PDT and to the battery snap...
am I doing something wrong?
cheers
Nothing wrong with that per se. But this standard wiring (http://www.madbeanpedals.com/tutorials/downloads/StandardWiring_MBP.pdf) only switches the battery with the input jack. And this is fine for most folks. Because who likes to unplug all boxes on a pedal board if you can simply switch off the power supply?
Psst. But I you really want to also switch the DC jack, simply change the wire going from the DC jack to the sleeve of the input jack so that it goes to the ring like the black wire of the battery clip.
hahaha, better to switch the bloody PS off!
but thanks for tips, very helpful for understanding the mechanisms!